$ E = \left[\begin{array}{rrr}1 & 2 & 3 \\ 4 & -2 & 3\end{array}\right]$ $ F = \left[\begin{array}{rr}3 & -2 \\ 3 & 0 \\ -2 & 5\end{array}\right]$ What is $ E F$ ?
Solution: Because $ E$ has dimensions $(2\times3)$ and $ F$ has dimensions $(3\times2)$ , the answer matrix will have dimensions $(2\times2)$ $ E F = \left[\begin{array}{rrr}{1} & {2} & {3} \\ {4} & {-2} & {3}\end{array}\right] \left[\begin{array}{rr}{3} & \color{#DF0030}{-2} \\ {3} & \color{#DF0030}{0} \\ {-2} & \color{#DF0030}{5}\end{array}\right] = \left[\begin{array}{rr}? & ? \\ ? & ?\end{array}\right] $ To find the element at any row $i$ , column $j$ of the answer matrix, multiply the elements in row $i$ of the first matrix, $ E$ , with the corresponding elements in column $j$ of the second matrix, $ F$ , and add the products together. So, to find the element at row 1, column 1 of the answer matrix, multiply the first element in ${\text{row }1}$ of $ E$ with the first element in ${\text{column }1}$ of $ F$ , then multiply the second element in ${\text{row }1}$ of $ E$ with the second element in ${\text{column }1}$ of $ F$ , and so on. Add the products together. $ \left[\begin{array}{rr}{1}\cdot{3}+{2}\cdot{3}+{3}\cdot{-2} & ? \\ ? & ?\end{array}\right] $ Likewise, to find the element at row 2, column 1 of the answer matrix, multiply the elements in ${\text{row }2}$ of $ E$ with the corresponding elements in ${\text{column }1}$ of $ F$ and add the products together. $ \left[\begin{array}{rr}{1}\cdot{3}+{2}\cdot{3}+{3}\cdot{-2} & ? \\ {4}\cdot{3}+{-2}\cdot{3}+{3}\cdot{-2} & ?\end{array}\right] $ Likewise, to find the element at row 1, column 2 of the answer matrix, multiply the elements in ${\text{row }1}$ of $ E$ with the corresponding elements in $\color{#DF0030}{\text{column }2}$ of $ F$ and add the products together. $ \left[\begin{array}{rr}{1}\cdot{3}+{2}\cdot{3}+{3}\cdot{-2} & {1}\cdot\color{#DF0030}{-2}+{2}\cdot\color{#DF0030}{0}+{3}\cdot\color{#DF0030}{5} \\ {4}\cdot{3}+{-2}\cdot{3}+{3}\cdot{-2} & ?\end{array}\right] $ Fill out the rest: $ \left[\begin{array}{rr}{1}\cdot{3}+{2}\cdot{3}+{3}\cdot{-2} & {1}\cdot\color{#DF0030}{-2}+{2}\cdot\color{#DF0030}{0}+{3}\cdot\color{#DF0030}{5} \\ {4}\cdot{3}+{-2}\cdot{3}+{3}\cdot{-2} & {4}\cdot\color{#DF0030}{-2}+{-2}\cdot\color{#DF0030}{0}+{3}\cdot\color{#DF0030}{5}\end{array}\right] $ After simplifying, we end up with: $ \left[\begin{array}{rr}3 & 13 \\ 0 & 7\end{array}\right] $